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Fie ABCD un trapez isoscel,AB║CD ,AB=24 cm si CD=9 cm .Daca M ∈ (AD) astfel incat [tex] \frac{DM}{AM} = \frac{3}{2} [/tex] ,iar MN║AB,N∈(BC),calculati segmentului [MN]

Răspuns :

Renatemambouko
DM/AM=3/2           
DM/(DM+AM)=DM/AD=3/(2+3)=3/5
MN intersectat cu DB={P}
DM/AD=MP/AB    3/5=MP/24    MP=24*3/5=72/5=14,4

ptr ca trapezul este isoscel putem scrie
CN/NB=3/2    (CN+NB)/NB=CB/NB=(3+2)/2=5/2    si NB/CB=2/5
NB/CB=NP/CD
2/5=NP/9
NP=2*9/5=18/5=3,6
MN=14/4+3,6=18 cm

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